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Author: Tinku Tara

Question-203750

Question Number 203750 by patrice last updated on 27/Jan/24 Answered by witcher3 last updated on 27/Jan/24 $$\frac{\mathrm{4}\left(\mathrm{k}+\mathrm{2}\right)−\mathrm{k}}{\mathrm{k}\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{4}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{1}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}−\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\mathrm{V}_{\mathrm{k}} −\mathrm{V}_{\mathrm{k}+\mathrm{2}} \\ $$$$\mathrm{s}_{\mathrm{n}}…

Question-203747

Question Number 203747 by patrice last updated on 27/Jan/24 Answered by esmaeil last updated on 27/Jan/24 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{\mathrm{1}+{cosx}}{dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}}{\mathrm{1}+{cosx}}{dx} \\ $$$${x}={u}\rightarrow{dx}={du} \\ $$$$\frac{{dx}}{\mathrm{1}+{cosx}}={dv}\rightarrow{v}={tan}\frac{{x}}{\mathrm{2}}…

Q-In-AB-C-AB-AC-b-and-BC-a-If-cos-A-a-2-b-2-then-find-the-value-of-b-a-

Question Number 203704 by mnjuly1970 last updated on 26/Jan/24 $$ \\ $$$$\:\:\:\:\:{Q}:\:{In}\:{A}\overset{\Delta} {{B}C}\::\:\:\left(\:{AB}={AC}={b}\:\right)\:{and}\:\:\left(\:{BC}={a}\:\right).\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{If}\:\:,\:\:{cos}\:\left({A}\:\right)=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:,\:\:\:\:{then}\:{find}\:{the}\:{value}\:{of}\::\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\frac{{b}}{{a}}\:}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by…

Question-203701

Question Number 203701 by Noorzai last updated on 26/Jan/24 Answered by mr W last updated on 26/Jan/24 $$\mathrm{0}<\frac{\mathrm{2}^{{x}} }{{x}!}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}×….×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×{x}}<\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \\ $$$$\mathrm{0}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} }{{x}!}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0}…